How do you solve #cos2x+3cosx-1=0#?

Question

17745 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-25T18:13:03

After you substitute #2cos^2(x) - 1# for #cos(2x)#, it becomes quadratic equation where #u = cos(x)#.

#2cos^2(x) - 1 + 3cos(x) - 1 = 0#

#2u^2 + 3u - 2 = 0#

#(u + 2)(2u - 1) = 0#

#u = -2# is extraneous, because #cos(x)# cannot equal -2

#cos(x) = 1/2# is the only possible root.

#x = cos^-1(1/2)#

For the first quadrant:

#x = pi/3# radians or #60°#

There is a root in the 4th quadrant, #5pi/3# and, of course, roots at multiples #2pi# for each.